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3 years ago

9

The following table lists the speed of sound in various materials. Use this table to answer the question.

2 answers:

defon3 years ago

5 0

Answer:

4.9sec

Explanation:

Lynna [10]3 years ago

4 0

Answer: Time taken by a sound impulse to travel through an aluminum rod 25 kilometers long is 4.9 seconds.

Solution:

Length of aluminium rod = 25 km = 25000 m (1 km=1000 m)

Time taken by sound to travel through aluminium = 5100 m/s (from the table given)

$speed=\frac{distance}{time}=\frac{25000 m}{\text{time taken}}$

time taken = $\frac{distance}{speed}=\frac{25000 m}{5100 m/s}=4.9019 seconds\approx 4.9 seconds$

Hence, the correct option is 4.9 seconds.

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zimovet [89]

Answer:

Your body cells use the oxygen you breathe to get energy from the food you eat. This process is called cellular respiration. During cellular respiration the cell uses oxygen to break down sugar. Breaking down sugar produces the energy your body needs.

Explanation:

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3 years ago

Stock naming please help

andre [41]

Answer:

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3 years ago

Measurements show that unknown compound X has the following composition: element mass % carbon 41.0% hydrogen 4.58% oxygen 54.6%

anastassius [24]

Answer:

CHO

Explanation:

Carbon = 41%, Hydrogen = 4.58%, oxygen = 54.6%

Step 1:

Divide through by their respective relative atomic masses

41/ 12, 4.58/1, 54.6/16

3.41 4.58 3.41

Step 2:

Divide by the lowest ratio:

3.41/3.41, 4.58/3.41, 3.41/3.41

1, 1, 1

Hence the empirical formula is CHO

8 0

3 years ago

6) A student measures out 96.21 g of sulfur for an experiment. How many moles of Sulfur are in this

barxatty [35]

Atomic mass of Sulfur = 32g

32g of Sulfur is one mole.

1g of Sulfur is $\frac{1}{32} moles$

96.21g of Sulfur is $\frac{96.21}{32} moles=> 3moles(appx)$

5 0

3 years ago

Read 2 more answers

Calculate the unit cell edge length for an 78 wt% Ag- 22 wt% Pd alloy. All of the palladium is in solid solution, and the crysta

BaLLatris [955]

Answer:

The unit cell edge length for the alloy is 0.405 nm

Explanation:

Given;

concentration of Ag, $C_{Ag}$ = 78 wt%

concentration of Pd, $C_P_d$ = 22 wt%

density of Ag = 10.49 g/cm³

density of Pd = 12.02 g/cm³

atomic weight of Ag, $A_A_g$ = 107.87 g/mol

atomic weight of iron, $A_P_d$ = 106.4 g/mol

Step 1: determine the average density of the alloy

$\rho _{Avg.} = \frac{100}{\frac{C_A_g}{\rho _A_g} + \frac{C__{Pd}}{\rho _{Pd}} }$

$\rho _{Avg.} = \frac{100}{\frac{78}{10.49} + \frac{22}{12.02} } = 10.79 \ g/cm^3$

Step 2: determine the average atomic weight of the alloy

$A _{Avg.} = \frac{100}{\frac{C_v}{A _A_g} + \frac{C__{Pd}}{A _{Pd}} }$

$A _{Avg.} = \frac{100}{\frac{78}{107.87} + \frac{22}{106.4} } = 107.54 \ g/mole$

Step 3: determine unit cell volume

$V_c=\frac{nA_{avg.}}{\rho _{avg.} N_a}$

for a FCC crystal structure, there are 4 atoms per unit cell; n = 4

$V_c=\frac{4*107.54}{ 10.79*6.022*10^{23}} = 6.620*10^{-23} \ cm^3/cell$

Step 4: determine the unit cell edge length

Vc = a³

$a = V_c{^\frac{1}{3} }\\\\a = (6.620*10^{-23}}){^\frac{1}{3}}\\\\a = 4.05 *10^{-8} \ cm$ = 0.405 nm

Therefore, the unit cell edge length for the alloy is 0.405 nm

7 0

3 years ago